\(\dfrac{x+10}{27}\text{=}\dfrac{x}{9}\)
cho biểu thức
P=(\(\dfrac{\text{x^3+3x}}{\text{x^3+3x^2+9x+27}}\)+\(\dfrac{\text{3}}{\text{x^2+9}}\)):(\(\dfrac{\text{1}}{\text{x-3}}\)-\(\dfrac{\text{6x}}{\text{x^3-3x^2+9x-27}}\))
rút gọn p
với x>0 thì P không nhận gt nào
Tìm cácgt của x để P nguyên
ĐKXĐ: \(x\ne\pm3\)
\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)
\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)
\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
Ý 2 mình k hiểu ý bạn lắm
\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)
\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)
\(\dfrac{16}{24}+\dfrac{10}{9}\text{×}\dfrac{27}{11}\) =
=
Rút gọn N= \(\dfrac{\text{2^19 x 27 ^3 - 15 . 4^9 . 9^4}}{\text{69 x 2^10 - 12^10}}\)
P = \(\dfrac{2.4.10+4.6.8+14.16.20}{3.6.15+6.9.12+21.24.30}\)
Giups mik trong thời gian ngắn nhất vs . cảm ơn
\(N=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot5\cdot2^{18}}{3^9\cdot2^9\cdot2^{10}-2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\cdot\left(2-5\right)}{3^9\cdot2^{19}\left(1-2\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{-3}{-5}=\dfrac{3}{10}\)
\(P=\dfrac{8\cdot10+8\cdot24+8\cdot560}{6\cdot45+6\cdot108+6\cdot120\cdot21}=\dfrac{8\left(10+24+560\right)}{6\left(45+108+120\cdot21\right)}=\dfrac{4}{3}\cdot\dfrac{2}{9}=\dfrac{8}{27}\)
Giải các phương trình sau:
1. \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2\text{x}-\dfrac{10-7\text{x}}{3}}{2}-x-1\)
2.\(\dfrac{3}{10}\left(1,2-x\right)-\dfrac{5+7x7}{4}=\dfrac{1}{20}\left(9\text{ }x+0,2\right)-\dfrac{12,5x+4,5}{5}\)
a,\(\dfrac{1}{7}\text{x}\dfrac{2}{7}+\dfrac{1}{7}\text{x}\dfrac{5}{7}+\dfrac{6}{7}\) b,\(\dfrac{6}{11}\text{x}\dfrac{4}{9}+\dfrac{6}{11}\text{x}\dfrac{7}{9}-\dfrac{6}{11}\text{x}\dfrac{2}{9}\)
c, \(\dfrac{4}{25}\text{x}\dfrac{5}{8}\text{x}\dfrac{25}{4}\text{x}24\)
`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)
\(=\dfrac{6}{11}.\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)
c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)
\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)
\(=1.15=15\)
Bài 1: Cho biểu thức \(A=\dfrac{x+2\text{√}x-10}{x-\text{√}x-6}-\dfrac{1}{\text{√}x+2}-\dfrac{\text{√}x-2}{\text{√}x-3}\) với x ≥ 0 và x ≠ 9
a) Rút gọn A
b) Tính giá trị của A khi x = 9-4√5
c) Tìm giá trị của x để A = \(\dfrac{1}{3}\)
Bài 15:
a)\(\dfrac{-2}{5}\)+\(\dfrac{4}{5}\) . x =\(\dfrac{3}{5}\)
b)\(\dfrac{-3}{7}\) - \(\dfrac{4}{7}\):x = -2
Bài 16
a) x - \(\dfrac{10}{3}\) = \(\dfrac{7}{15}\) . \(\dfrac{3}{5}\)
b) x + \(\dfrac{3}{22}\)= \(\dfrac{27}{121}\) . \(\dfrac{11}{9}\)
c) \(\dfrac{8}{23}\) . \(\dfrac{48}{24}\) - x = \(\dfrac{1}{3}\)
d) 1 - x = \(\dfrac{49}{65}\).\(\dfrac{5}{7}\)
Bài 17: tìm x
a) \(\dfrac{62}{7}\) . x = \(\dfrac{29}{9}\): \(\dfrac{3}{56}\)
b) \(\dfrac{1}{5}\) : x=\(\dfrac{1}{5}\)+\(\dfrac{1}{7}\)
bài 18:
a)\(\dfrac{2}{5}\)+\(\dfrac{3}{4}\): x =\(\dfrac{-1}{2}\)
b)\(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\)x + \(\dfrac{3}{5}\)x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\).x-x = \(\dfrac{-9}{14}\)
bài 19: tính
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+ \(\dfrac{1}{2018.2019}\)
bài 20:tìm x
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{2008}{2009}\)
bài 21: tìm x
\(\dfrac{x+1}{99}\)+\(\dfrac{x+2}{98}\)\(\dfrac{x+3}{97}\)\(\dfrac{x+4}{96}\)=-4
bài 22 : so sánh các phân số sau:
a) \(\dfrac{-1}{5}\)+\(\dfrac{4}{-5}\)và 1
b) \(\dfrac{3}{5}\) và \(\dfrac{2}{3}\)+\(\dfrac{-1}{5}\)
c)\(\dfrac{3}{2}\)+\(\dfrac{-4}{3}\) và \(\dfrac{1}{10}\)+\(\dfrac{-4}{5}\)
d) \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{1}{6}\) và 2
tìm GTNN
C=\(\dfrac{x^6+27}{\text{x}^4-3x^3+6x^2-9x+9}\)
\(C=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^4+3x^2-3x^3-9x+3x^2+9}=\dfrac{\left(x^2+3\right)\left(x^4+6x^2+9-9x^2\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}\\ C=\dfrac{\left(x^2+3\right)^2-9x^2}{x^2-3x+3}=\dfrac{\left(x^2-3x+3\right)\left(x^2+3x+3\right)}{x^2-3x+3}\\ C=x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
Câu 1 : -\(\sqrt{9}+\sqrt{0,25=}\)
A. 3,5 B.-3,5 C.2,5 D-2,5
Câu 2 :\(\sqrt{\dfrac{9}{6}-\sqrt{ }6^2}=\)
A-\(\dfrac{21}{4}\) B\(\dfrac{21}{4}\) C-\(\dfrac{27}{4}\) D\(\dfrac{27}{4}\)
Câu 3 : 2,5 . x - 3,35 = -10 nên:
A.x=2,65 B.x= -2,66 C.x=2,67 D.x= 2,68
Câu 4 :Mai và Lan cùng nhau làm mứt dừa theo công thức cứ 2 kg vừa thì cần 3 kg đường . Hỏi hai bạn làm mứt từ 2,5 kg dừa thì cần bao nhiêu kg đường?
A .3,5 B.3,6 C.3,75 D.3,8
Câu 5 :Nếu x và y là hai đại lượng tỉ lệ nghịch và x=4, y=42 thì hệ số tỉ lệ của y đối với x là:
A.168 B.178 C.169 D.160
Câu 6 : Hàm số y = f(x) = 4 . x -\(\dfrac{4}{3}\). Tính f (\(\dfrac{1}{3}\)) là :
A.\(\dfrac{1}{3}\) B.0 C.\(\dfrac{4}{3}\) D.\(\dfrac{5}{3}\)
Câu 7 : Cho hàm số y = f(x) = x\(^2\) - 5 . Khi đó :
A.f(1)=4 B.f(-2) = -9 C.f(1) >f(-1) D.f(2)= f(-2)
Mn giúp em với ^^